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How much horsepower to weight is required for a 1:1 thrust ratio?
I did a math experiment today, but wanted to verify the results.
I wanted to figure out how much horsepower is needed for an aircraft to hover in midair with no wings/lifting body (1:1 thrust/weight).
So to start I found that a unit of horsepower is the amount of work required to move 550 pounds 1 foot every second.
Then I wanted to find out how many foot/pounds of force gravity places on 1 pound of material. With a static acceleration of 9.8m/s (32f/s), it seems that 1 pound has 32 foot pounds of gravity acting on it.
Now for my equation:
Hp = 550 (f/p)
Gravity = 32 (f/p)
Hp / Gravity = 17.1875

So it seems 1 hp can hold 17 pounds in the air, but to incorporate propeller inefficiency, I just used 70% of that. This leaves me with 1 hp being able to vertically lift 12 pounds when accounting for the inefficiency of a propeller.
Is this a relatively correct assumption? I know thrust changes as the aircraft accelerates, but in a hovering state the acceleration is 0 f/s as a whole. So will these numbers apply?
Edit: I initially used a specific ultralight to prove how high the numbers were, but I used wrong data from the spec sheet. So I removed that reference from my question.

You seem to be confusing acceleration, power and thrust, and then running with it into increasingly more erroneous calculations. What you should be asking is how much of the engine's rated power can be converted into thrust by the propeller. Thrust is a force, which you can compare to gravity. Power, on the other hand is energy per unit of time.

Don't you need the gross weight from the spec sheet, which is ~1000lbs? According to Wikipedia, the engine alone weighs 110lbs, so it seems your estimate that the whole aircraft weights 115lbs is too low.

@NuclearWang oh wow... yeah that needs rectifying. Maybe the math isn't as off as I thought

@AEhere Isn't thrust the amount of force an engine can generate in it's given environment? Since the engine acceleration is 0 compared to the air it's moving through (assuming we are hovering) then isn't the only thing left in covering power to force friction & inefficiencies?

@YAHsaves no, please see this basic explanation. An even simpler check you can perform is to look at the units: power and force use different physical units, even if you ignore inefficiency, you are clearly missing some other units to make your $T=k \cdot P$ expression dimensionally correct, i.e. $k$ needs to have units as well, unlike a dimensionless friction coefficient.

1 horsepower = 1 horse worth of thrust, ergo horses can fly/hover. =)

Many top-end R/C model aircraft can thrust hover -- some have been built with control surfaces not normally found on aircraft to facilitate this (roll surfaces directly behind the propeller, for instance).  A .60 cu. in. glow engine (around 3 hp) spinning a correctly sized propeller to let the engine develop peak power can do vertical pull-ups (fly upward out of a hover with around 2G acceleration).  This is close to that 3 lb thrust per HP figure, so it's robust and scales well.

@EricHauenstein one Clydesdale can exert a peak of roughly 17 horsepower.  So they should accelerate vertically like a rocket :)

Acceleration is distance per time squared, not distance per time. A pound force is a pound not 32 pounds. A pound mass weighs one pound in standard gravity. Units have to be right to do any dimensional analysis.

Horsepower cannot be converted into thrust without knowing the speed at which this is done. In your case this is zero, so the special equation for static thrust applies. With $T$ for thrust, $P$ for power, $\rho$ for air density and $d_P$ and $\eta_P$ the propeller diameter and efficiency, respectively, this is $$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{P}^2\cdot\pi\cdot d_P^2\cdot\frac{\rho}{2}}$$
With this equation you cannot simply say how many HP are required to lift that many pounds; instead, you need to add the propeller geometry. If we assume a very efficient propeller with a large diameter, we can combine blade size and mass into disk loading – here 8.5 kg/m² or 41.5 lbs/ft² is a good value. Now we can write, using standard atmospheric density at sea level:
$$\frac{T_0^2}{P^2\cdot\eta_{P}^2} = \frac{2\cdot1.225\,\frac{\text{kg}}{\text{m}^3}}{8.5\,\frac{\text{kg}}{\text{m}^2}\cdot g} = 0.0294\,\frac{\text{s}^2}{\text{m}^2}$$
assuming for simplicity a 1 m propeller diameter. If we now use 70% for propeller efficiency, as you did in the question, the power to mass ratio becomes 8.33 Watt power per N of weight force or 81.7 Watt per kg. Converted into funny units this is 0.05 HP per pound of mass. 1 HP can hold 20 lbs in the air, but that will not be enough to even spin up the propeller unless it can adjust blade pitch. For practical usability I recommend to double that number. If the propeller diameter shrinks and disk loading increases, efficiency will drop and the power required will rise.

Seeing the numbers laid out like this helps me understand where all the "wasted" energy is actually going. If I understand it right my initial estimate of "12 pounds per hp" wasn't too far off, but that's assuming perfect air pressure and a very large prop. I see why real world conditions almost never meet that, considering the number of variables accounted for.

DIYDrones.com listed directly measured electric motor/prop data at around 1 kg thrust/100 watts Power, fairly linear through RPM range.  Yours numbers (even with a great variety of prop/rotor design), are close.

For practical usability you really should be running at around half or 3/4 power to generate hover thrust otherwise you would not have enough power budget to maintain altitude in case of disturbance. So yeah, I strongly agree with doubling the power. For RC planes you wouldn't even attempt to hover unless your plane has 2:1 thrust-to-weight

"For R/C planes you wouldn't even attempt to hover unless your plane has 2:1 thrust to weight"  I don't think even the Saturn 5 was that good.   But at optimal AOA (and adequate Reynolds number) Thrust to horsepower = lift to drag. I wonder how good that V-22 Osprey prop really is.

did you really mean to say "assuming for simplicity a 1 m² propeller diameter"?  Diameter is a distance, but m² is an area.

@Michael: Better now?

@RobertDiGiovanni It's actually easier to have a higher thrust to weight ratio in a R/C plane than it is a full-size one. Just to compare apples to apples, the Saturn V at launch has a TWR of about 1.2, while an Estes Bandito (a beginner model rocket) has a TWR of about 18 (Note: I didn't forget a period there, that's an eighteen, not a one point eight).

Yes, much better.  My initial thought was that you actually meant an area of a square meter, which now is obviously not the case.

@Michael: As usual, if there is a suspicion of malice or stupidity, stupidity is the most likely explanation. I copied and pasted a lot when writing the answer and have no better explanation how this happened. Thank you for spotting this!

@HiddenWindshield: I guess the Estes Bandito doesn't have svivelling rocket nozzles, so it needs those 18 in order to accelerate quickly enough without toppling over bevor the fins begin to work properly.

Thrust from a propeller aircraft will vary with airspeed, propeller efficiency, density altitude, etc.
For a very rough guideline you can use 1 hp to equal approximately 3 lbs of thrust. So theoretically a 3,000 lb aircraft could hover if it had about 1,000 horsepower.
The Lockheed XVF weighed approximately 15,000 lbs and could hover using about 5,000 hp.

Here is a link to some detailed testing regarding propeller thrust: Static Thrust Measurement for Propeller-driven Light Aircraft

Does that imply that propeller aircraft have terrible efficiencies? Total energy should be preserved, so if the propeller is only accelerating 1/17th the mass of air downward that it could be based on power input, where does the other 16/17th of the power go? Is it lost to circular momentum friction and vibrations?

I have edited my answer to correct wrong formation. I have also added some interesting photos and links.

max takeoff weight: 1,370 lb (635 kg) & -A2C flat 4 piston engine, 124 hp (93 kW) so 10 pounds per horsepower, is easily seen in production machines

@James Jenkins its pounds of THRUST per horsepower, not aircraft weight. That ratio is commonly thrust = 25-30% of aircraft weight (the miracle of wings)  But we are also looking at helicopter rotors.

In order to hover in the air, you have to accelerate a mass of air downwards. Usually, this is done with a rotor, and the larger the rotor, the less the power you'd need to hover. The general expression (in theory...)for the power required, based upon momentum considerations, is:
$P_{required}=T\sqrt{\frac{T}{2\rho\cdot A}}$
where T is the thrust (should be the same as the weight, in order to hover...), A is the rotor disk area, and $\rho$ is the air density.
Using SI units, thrust/weight should be in newtons, A in square meters, and $\rho$ is, at sea level, 1,23 kg/m3.

Seeing this equation really helps. I guess we are losing a lot of power due to the environment we are working with. However I don't know where to plug "rho" into the question. Is "rho" the "2p" or how do I apply it?

Yes, rho is the 'p'...For example, if your craft has a mass of 500 kg, that would mean a weight/thrust of 4900 newton. Assuming your rotor has 8m diameter, the disk area will be 50,24 m2. Inserting those figures in the expression, the power will be 30,9 kW. In practice, you'd need 50% more at least, since reality is usually hard...

A real world example for comparison: Robinson R22 ha MTOWof 622 kg, rotor diameter of 7.7m and max power of 124hp. Max hover altitude is 8000 (~MTOW) and assuming 3% power loss /1000ft, I'd say it takes about 95hp to hover R22 at 622kg (when not in ground effect). Feel free to correct if I made a mistake :) It's worth noting that some of that 95 hp (or what ever if I didn't get it right) is "lost" in couteracting the main rotor torque with the tail rotor.

Or, T is proportional to W ^ (3/2) (sorry no equation markup). And for a lifting aircraft in level flight, this happens for minimum power at a velocity which can be derived.

You've made quite a lot of mistakes in your math, some less severe and some more severe; you may want to think about these.
Notation mistakes
These mistakes are not so severe; we can still tell what you're talking about.
The unit of energy you're talking about is called the "foot pound" or the "foot-pound," not the "foot/pound." A foot pound is a foot multiplied by a pound. A slash in a unit indicates division, so a "foot/pound" is a foot per pound, or, in other words, a foot divided by a pound.
You can abbreviate "foot pound" as "ft lbf," but certainly not as "f/p."
Note that in the phrase "food pound," the word "pound" means "pound-force."
Unit mistakes
These mistakes are more severe; as a result of these mistakes, your calculations are completely wrong.
Let's go paragraph by paragraph.

So to start I found that a unit of horsepower is the amount of work required to move 550 pounds 1 foot every second.

That's correct.

Then I wanted to find out how many foot/pounds of force gravity places on 1 pound of material.

There's no such thing as "foot pounds of force." Foot pounds are a unit of energy, not force. You can ask how many foot pounds of energy, or how many pounds-force of force (or just "pounds of force" for short), but not how many foot pounds of force.

With a static acceleration of 9.8m/s (32f/s), it seems that 1 pound has 32 foot pounds of gravity acting on it.

There are two mistakes here. The acceleration due to gravity is $9.8\ \mathrm{m}/\mathrm{s}^2$ (meters per second per second) or $32\ \mathrm{ft}/\mathrm{s}^2$ (32 feet per second per second), not $9.8\ \mathrm{m}/\mathrm{s}$ (meters per second) or $32\ \mathrm{ft}/\mathrm{s}$ (32 feet per second).
The second mistake is that you've changed feet per second into foot pounds. A foot per second is not the same thing as a foot pound.
There are two ways of saying this correctly:

1 pound-mass has 1 pound-force of gravity acting on it.
1 pound-mass has 32 pound-mass feet per second per second of gravity acting on it.

Now for my equation:
Hp = 550 (f/p)
Gravity = 32 (f/p)
Hp / Gravity = 17.1875

The numbers are correct, but since the units on the first two quantities are wrong, the units on the last quantity are wrong, too.
With the correct units, the calculation is:
$$\text{Power} = 550\ \mathrm{ft\ lbf}/\mathrm{s}\\
\text{Gravity} = 32\ \mathrm{ft}/\mathrm{s}^2\\
\begin{align}\text{Power} / \text{Gravity} &= 17.19\ \mathrm{lbf\ s}\\
&= 17.19\ \mathrm{slug\ ft}/s\\
&= 550\ \mathrm{lb_m\ ft}/s\end{align}$$
(No need for more than four significant digits of precision.)
See the definition of a "slug" on Wikipedia.
So the correct interpretation of this result is that one horsepower is enough to raise 17.19 slugs (equivalently, 550 pounds) one foot per second; or 1 slug 17.19 feet per second; or 1 pound-mass 550 feet per second. This calculation is good for raising the mass using pulleys or something; it doesn't work for aircraft.

So it seems 1 hp can hold 17 pounds in the air

Here you've made yet another mistake: the result of your last calculation was the unitless number 17 (and it was supposed to be 17 slug feet per second), but you changed that to 1 pound.
I don't think you made any other mistakes besides these.

I think the fundamental mistake is using the archaic measurements of the ancient British. Is a slug lighter than a snail?

Work is force × distance, and power is work per time.  If distance travelled is zero (object is in motionless hover), work performed and power applied also are zero.

If that were the entire story, you'd be able to shut down a helicopter engine while hovering and stay in the air. So this is an oversimplification.


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Would putting several small propellers on a wing be efficient?
I saw this video that NASA posted, which presents the idea of putting several small propellers on a wing:

They claim that by doing this you can force air to blow over the wing more quickly. This would make the wings act as if the aircraft was flying faster than it really was, resulting in shorter takeoffs and more control on landings.
However while watching the video I kept thinking, wouldn't the turbulence and slipstreams caused by the propellers actually make the wings perform worse? Or at least neutralize the proposed added lift and control?
Having no fresh airflow over the wings seems like a problem. Have any aircraft ever tried this concept before with successful results?

AeroVironmnet, working with NASA, has flown 5 UAVs like this.

It would provide pretty good STOL performance at low speed because there is a significant increase in lift with the wing in the slipstream of a propeller, especially with flaps down.  One of the biggest differences between a jet and a turboprop while flying a landing approach is that power settings have a direct effect on sink rate with a turboprop where a quarter or a third of the wing span is in the propeller slipstream; not so much on a jet.
Then when you speed up, the inefficiency of all those draggy nacelles and relatively inefficient smaller propellers, the weight, the maintenance nightmare, the fuel burn.  Yikes.  That's why it's not done.
There's a reason that airplanes are all built to a few near-universal configurations.  They are the arrangements that hit a sweet spot, in the balance of compromises game, for the airplane's mission, and they are timeless until some major new paradigm comes along.  That's not one of them.

Brilliant answer thank you! That was something along the lines of what I expected. So would it be correct to assume that all these propellers having a wing in their slip stream would cause additional thrust lost to exchange for lift in an already fuel thirsty design.

Well if you had a 2000 HP engine on the wing replaced with 6 333 HP engines, the airplane could probably fly quite a bit slower, with power on, than with one engine on the wing, especially if there was a full span flap, but just about every other parameter would be much worse.   The other thing is there would be a huge difference between minimum flying speed power on and power off.  You'd slow to some speed with approach power and if you went to idle it would plummet.  Regular turboprops already do that to some degree.

Re "There's a reason..." isn't a big part of the reason that it's much easier to build and maintain (and control) a few large IC engines than a multitude of smaller ones?  So if you have workable electric engines, you can use many of them, as with NASA's X-57: https://www.nasa.gov/centers/armstrong/news/FactSheets/FS-109.html

This design is not intended for combustion engines like you assume, but for electric aircraft, and indeed this has been done 25 year ago. The many smaller motors reduce the need for spanwise load carrying and provide yaw control without a rudder. Another factor is increased redundancy for UAVs that remain flying for months at a time.

Even if they are electric motors, you have somewhat the same problem.  The thrust per watt would be low compared to a single large propeller on each wing, because of many small inefficient propellers, so there is a benefit but it comes with a host of negatives.  If I was building an electric transport aircraft I would still want the fewest engines with the largest propellers I could get.  For a UAV, u have a good point but there is still a huge price for that redundancy, although for a specialized use like that it could be worth paying.  Depends on the  objectives.

@John K: NASA seems to have a different opinion, per the link I posted above.

The F-4 Phantom (early models)  had a system called Boundary Layer Control (BLC). that, when the flaps were down (like for landing), release high pressure high velocity air diverted from the last stage of the engine compressor, out of multiple small holes in the leading edge of the wing. It basically "reenergized the boundary layer air flow over the top of the wing, and effectively did exactly what your question postulates. As a result, landing speeds were reduced by 10-15 knots. Prop wash does the same thing. Problem with your idea is that prop wash is localized to just behind the prop, not spread out over the entire wing. Multiple propellers require multiple engines, or some mechanical system to distribute the power from one engine to multiple propellers (weight!).
A perhaps better solution, (I'd have to do the engineering!), would be to use multiple small discharge holes in the leading edge of the wing, as the F-4 did, hooked up to a BLC system using high pressure, high velocity air from an engine driven compressor.
But I suspect that the benefits (a slight reduction in landing speeds) might not be sufficient to mitigate the additional weight, complexity, cost and problems of such a system.  Indeed, the BLC system in the F-4 was eventually ripped out due to the problems associated with system failures.

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Can lift be equated to how much air is moved?
The explanation for lift that I find easiest to visualize is that air is deflected downward by the wing. This downward action causes an equal and opposite reaction of an upward force against the plane, as in accordance with Newton's third law.
As you can see from the image below the red lines both end lower than they started.

Air at sea level weighs 1.293 kg per cubic meter. The acceleration due to gravity is 9.8m/s².
That comes up to 9.8N of force being needed to keep a 1kg aircraft in the sky. With perfect efficiency this equals 9.8 watts.
If trying to hover VTOL aircraft, can you multiply how much air a propeller can move, by how fast it moves it to predict lift force? Since force is mass x acceleration?
Or do fluid dynamics work differently as the air exiting the propeller will spread out instead of all going straight down? If that is the case can using a duct with a properly fitted bell nozzle solve this problem? Or am I thinking into realms that I shouldn't be?

First off, your calculation for 9.8 W does not seem to be based on any actual physics. Check the units! I can't tell if you did $P=F\cdot m$ or $P=\frac{F}{m}$, but the result is either the nonsensical quantity of joule-kilogram per meter, or acceleration respectively.
A correct calculation based on keeping something aloft by propelling air downwards would be as follows. To keep something aloft, you need to expel mass downwards at a certain speed. The force to keep the object aloft is $$F=m_{object}g$$ The force generated by downwards momentum transfer is $$F=\dot{m}v$$ with $\dot{m}$ indicating mass flow (kilogram per second) of the air (not the mass of the object). The energy flow (power) required to  impart this momentum on the airflow is $$ P=\frac{1}{2}\dot{m}v^2$$ Here we can draw an important conclusion. The power requirement is arbitrarily small, by increasing the mass flow and decreasing the downwards velocity. Immediately, you can see why helicopters have big rotors and airplane wings are so large: they want to affect as much air mass as possible for increased efficiency. Even flying faster increases lift efficiency (reduced induced drag) by affecting more air per unit of time. Of course, at some point other sources of drag will dominate.

The above calculations assume that force is purely generated by creating a downwards mass flow. This is not exact reality. For example, put your aircraft on the ground, and it does not sink through it even though no mass flow is created. When flying close to the ground, some of the airflow downwards creates a pressure due to hitting the ground below, which helps you keep aloft as well (this is called the ground effect). Other than that, viscosity plays a small role by generating a force opposite the downwards air movement (although viscosity is required for lift generation with an airfoil). However, to answer the main question: equating lift by how much air is moved is a very good approximation.

Let's finally address your idea of using a Bell nozzle. This is a nozzle that is used on supersonic propellants (the 'throat' of the nozzle marks the transition from subsonic to supersonic), and is used to increase the speed as much as possible. This is very inefficient in terms of energy, but since rockets need to carry all the propellant mass (whereas an airplane gets 'free' mass flow by travelling through air), and need extra propellant to carry that propellant up, speed wins instead of mass, and rocket engines are optimised for nozzle velocity.

+1 Excellent answer!

@PeterKämpf Thank you. Do you perhaps know how much viscosity practically influences lift? (outside maintaining the Kutta condition)

@Sanchises The math I was trying to do when I came up with 9.8w was force in newtons  equals mass(kg) * acceleration(m/s).  Since one newton equals 1 watt I thought the conversion would work out. Your answer though was very good and I understand this a lot better thank you!

@YAHsaves Whoever told you one newton equals one watt was wrong. Power is force times velocity. One watt equals one newton-meter per second. Always check your units!

To answer your question from July 22 (sorry to have missed it!): Viscosity reduces the lift curve slope and limits maximum lift. Think of the boundary layer as an additional part of the wing which changes the shape of the airfoil. On the suction side this part becomes thicker as air slows down past the suction peak, so the angle of attack of the airfoil + boundary layer is lower than the geometric angle of attack of the naked airfoil. This becomes really dramatic when the flow separates from the airfoil.

@PeterKämpf No problem, I'm surprised you still found it. And also, thanks for the clear explanation. I suppose macroscopic effects are negligible then? (I'm thinking perhaps that spanwise flow is reduced ever so slightly, but given how long the downwash persists this effect should be minimal)

Spanwise flow in the boundary layer is greatly affected by viscosity and when the boundary layer grows to macroscopic levels (think local separation), it is no longer negligible for the outer flow.

@Sanchises " The power requirement is arbitrarily small, by increasing the mass flow and decreasing the downwards velocity" Because to double mass flow you double need double power and to double velocity you need 4 times power?

The point is not that the air ends up lower when leaving the wing, but that an impulse has been imparted on it such that it continues to move downwards behind the wing.
Proof: Read this answer on wake formation, and watch the animated GIF in this answer.
The next question is: How much air is involved?
That is much harder to pin down precisely because the boundaries of the air that is influenced by the wing are theoretically at infinity. But there is a practical approximation that is very similar to Froude's theory for propellers. Just take what flows through a circle with the diameter of the wingspan. This answer uses this idea to calculate induced drag, and it reaches the correct result.
So there you have it: Yes, lift can be equated to how much air is accelerated.

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How can this aircraft be stable/maneuverable?
The new wave of electric VTOL aircraft concepts (Joby Aviation, Zee, Lilium, Airbus A$^3$, etc...) has produced some audacious unconventional designs.
Now my question, is how Lilium's concept can be stable and/or maneuverable, given the following configuration:

What strikes me is a lack of a surface that creates a force to counteract the momentum generated by the distance between the wing's lift and the gravitational force.
And although control around the yaw axis seems possible with differential thrust, I wonder how the aircraft is controlled around the pitch axis.
From the forward part of the fuselage some canard-like surfaces with integrated propellers are deployed to provide hovering capabilities, however it seems these are supposed to be retracted during forward flight.

Some might suggest the pictures are pure artwork, but Lilium and its investors appear to believe in the concept:

10m€ funding from London-based venture capital firm

Those images represent art, not engineering.

Must be an anti-gravity system. I don't think the wings would generate any lift with the motors in the way of the airstream.

@JonathanWalters At first glance, that's what I would think too. However, Lilium claims to be developing exactly this aircraft. Someone appears to believe in this: https://medium.com/lilium-aviation/10million-funding-b49addcdfa71#.udrqmr9nv

Well then, I would love to know the expected operating limitations for turbulence and wind gusts. I have trouble imagining such a design could handle conditions such as 20 gusting to 39, orographic turbulence, and low level wind shear very well. A calm-weather-only design?

Could be body of the aircraft be generating lift?

I was going to say it can't glide, but then I read it has a whole-airframe parachute, so I stand corrected.

I don't understand the question.

While I don't think the magnitude would be enough to solve the problem, if all of the weight of the motors and batteries were at the trailing edge of the wing, the CG might be further aft than it intuitively appears.  But probably not enough further.  It would seem like you could move the wing forward a fair amount while still maintaining good downward visibility, question is if it would be enough visibility for a vertical hover landing, or only an extremely steep descent.

http://www.esa.int/var/esa/storage/images/esa_multimedia/images/2016/05/ducted_fans_on_wings/15974606-2-eng-GB/Ducted_fans_on_wings.jpg

I imagine that that beautiful thing will fly like a shuttlecock.

Some clips of a test flight were published in the media today, if that helps at all.

@RobieBasak just as fake as everything else about the thing. Note how the 'retractable forward surfaces' could never retract and leave room for a pilot's legs, or even at all because they together are wider than the room they have to retract into...

That test flight is clearly a scale model drone and it never retracts its forward "hovering" jets. It looks from the t if it did retract those it would nose down and fall quickly to its destruction. @Notts90 It does look like it might be designed as a lifting body.

What strikes me is a lack of a surface that creates a force to
  counteract the moment generated by the distance between the wing's
  lift and the gravity force.

You are not alone. Believe it or not, Lilium contacted me for advice, and I asked them basically the same thing. They never responded.
I agree with you, this thing will never fly and investors will get burned.
My advice to them would be to sweep the wings forward. This allows to keep the fuselage basically unchanged - you don't want the main spar to go where the pilot sits. But given the small wings this needs substantial sweep.

Update to this answer, February 2020:
Now they have beefed up their canard and skipped retractability, but all they can show is hovering flight for a few minutes at most. Sill no aerodynamic flight. The 300 km range and hour long flight time still are impossible with near-term technology. My verdict stands: Investors will get burned.

It's surprising that they would get to the full scale model without running wind tunnel tests or even modeling it in X-Plane to determine its characteristics. I'm guessing that this design would only work in VTOL configuration or where the front motors can act as a canard.

Their development page shows several flight tests, but all appear to have a forward ducted fan in the position where the circles are in the picture above:
http://lilium-aviation.com/development.html

"Lilium contacted me for advice..." -- we all knew Peter was good. I don't know that many of us knew he was that good!

I'd only ask whether those "wings" are indeed wings, or just streamlined mounting pylons for the electric motors.  I also wonder why the artwork isn't showing the extension cord, 'cause there's not room for enough batteries for more than a few minutes of flight :-)

Maybe there's a few tons of batteries in the tail. /eyeroll

@SteveS so it's a multicopter with some things sticking out of it that look like wings to make the occupants feel better.

Their 2:1 scale demonstrator model is definitely flying as a multicopter, there's not enough forward motion.  It's not as heavy as the 300kg it should weigh either (as a 2:1 model).  I think the only viable strategy for them is to use the investment capital to first build a time machine, which they could then use to steal Graphene engineering technologies and Graphene lithium-ion batteries from the future.

@Adrian Thompson Phillips: I don't think even graphene lithium-ion batteries are going to provide enough energy.  They need to go sideways into the Star Trek universe and bring back a bunch of dilithium crystals :-)

"this thing will never fly and investors will get burned": it apparently does fly, according to some short clips of a test flight published in the media today. Looks like it is using downwards pointing fans rather than using the wings for lift.

@RobieBasak: Thanks for the link. What you see is something that uses its wings only to place propellers, but not to create lift, as you rightly say. It is flying slowly and the movie even loops back and forth to create the impression of a longer flying time. I do not see a reason to reverse my judgement.

Here's a better video of their first drone flight

@Bergi: I wouldn't exactly call this flying - hovering yes, but not flying. For now the wings are purely for keeping the inefficient ducted props in place.

@SteveS I've seen many a kickstarter or similar website designed to lure venture capitalists and other investors with grandiose claims "supported" by videos and pictures that are clearly fakes to anyone with a bit of knowledge about physics or another related science exceeding primary school levels. This appear from the looks of it to be no different.

@m2as3registeredservicesohmone: They have hovered something that looks quite different to what it looked when I wrote this answer.

I guess buying an airline is not the only way to become  millionaire in aviation business...

It just goes to show one of my fundamental rules about aerospace: always beware of the Silicon Valley whiz kid claiming to be able to do what others have struggled to make for decades.

@PeterKämpf https://lilium-aviation.com/files/redaktion/refresh_feb2021/investors/Lilium_7-Seater_Paper.pdf a detailed performance assessment from Lilium.

@slitvinov What is needed is outside company engineers saying it will work, not based on the people asking for the money. If they want a vehicle that uses lift then that needs to be inherent in he design. We already know that blowing air at the ground really hard will get something off the ground. That isn't new technology. My issue with the pop out canard design is there is no place inboard to hide both of those units and then push them out, lock them into place to use them without being really heavy. Where do you put the the lever with nearly no space between the units when they are inboard?

This looks a lot like many other similar technology projects which are really more designed to generate a fast buck by farming  grants and crowdfunding than with any realistic expectation of ever producing a working product.
Given that there is no technical detail apart from concept art and there are no remotely similar designs which do fly the question of how it might fly is purely speculative and there is no meaningful engineering data to comment on.

Looks like they've noticed their stability issues and are trying to update the design. Permanent fixed wing forward now and winglets the rear wings.

What wonderful progress. What will they think of next, a vertical tail?

Yes, T, with another set of ducted fans on it.

Unless the body has some significant lifting properties, the nose will sink and the plane will not fly (without the canards extended).
In a no power situation, without a vertical stabilizer and some fore/aft lifting balance, this plane would be a lawn dart.

It probably cant.
This wing configuration is designed for extreme performance, as in, its extremely unstable. Military planes are often unstable, which makes them turn faster.
Civilian planes are supposed to be stable, making them turn slow, comfortable, and safe.
Creating an unsafe air transport that is very uncomfortable and difficult to fly so to create dogfight capability for a civilian craft is either A: Madness, or B: attempt to make it look cool at the cost of everything else.
So its a scam for investors.
One possibility is that the top of the engine housing is designed to work as an elevator, but the question once again is "why?".

Consider the third picture: look at the chord of wing and compare it to the attitude of the nose (angle of incidence). It is clear that the nose will point fairly high,  with a substantial amount of air hitting the belly of the aircraft, thus moving the centre of pressure forward, possibly close enough to the centre of mass.
There is an angle of incidence that neutralises the rotation around the pitch axis, although this does not result in a stable equilibrium condition. Some active control is required to avoid to flip the plane either downwards of backwards.

This is a comment (trying to convince myself), not an answer. -1

There is no way the nose can be very light since it's supposed to seat two people.

The lift, if there is any, could only emerge at high speed. Presumably you can get there from low speed regime without flipping over by pointing the front thrustors downward. But even then, it is going to be unstable.

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text:
Can a certified mechanic ground my airplane?
The courtesy (under-wing) lighting and the rear seat overhead map light were on the same switch. I didn't like having the bright outside light on while my passenger read his magazine, so I separated the lights to two different switches.  What will happen when this is discovered during my next annual? And can a certified mechanic ground my airplane for it?

When you say you separated the switch to two different switches, was the second switch in use for something else, or was it a blank switch?

Welcome to aviation.SE! If I understand correctly, you have two questions here: does a mechanic have the authority to ground your aircraft; and does your wiring change require a mechanic's approval? Is that accurate?

Is it worth differentiating between taking action to "ground" your airplane vs refusing to approve by not signing off the annual inspection?  Because in my mind they are distinctly different things.

@Pondlife My guess is that any wiring change not in accordance with the official aircraft manuals should never have the approval of a qualified mechanic and therefore lead to the aircraft failing its annual until corrected/reverted.

@jwenting Yes, that makes sense; I've mentioned it at the end of my answer.

The short answer is that a mechanic can't ground a plane. There are lots of discussions online about this, and the typical scenario seems to be that a mechanic "refuses to sign off an annual" or something similar. (There are other scenarios too, where the mechanic refuses to return the aircraft and/or logbooks to the owner because of a dispute over payment or work performed, but that's a legal issue, not an aviation one.)

AOPA has two useful articles (here, here) on this topic, written by Mike Busch (a well known A&P and author). He says:

Who can ground an aircraft?
FAR 91.7 forbids a PIC from flying an aircraft that is unairworthy,
although FAR 21.197 (Special Flight Permits) allows the FAA to grant
special dispensation to fly an unairworthy aircraft (usually for
repositioning purposes). There is no FAR that empowers a mechanic to
ground an aircraft. Mechanics are not the safety police.

When it comes to signing off on maintenance, there are two cases: required inspections (e.g. annuals), and general maintenance (e.g. an oil change). Required inspections are governed by 14 CFR 43.11.
43.11(a)(4) says:

[...] if the aircraft is found to be airworthy and approved for return
to service, the following or a similarly worded statement—“I certify
that this aircraft has been inspected in accordance with (insert type)
inspection and was determined to be in airworthy condition.”

43.11(a)(5) says:

[...] if the aircraft is not approved for return to service because of
needed maintenance, noncompliance with applicable specifications,
airworthiness directives, or other approved data, the following or a
similarly worded statement—“I certify that this aircraft has been
inspected in accordance with (insert type) inspection and a list of
discrepancies and unairworthy items dated (date) has been provided for
the aircraft owner or operator.”

In other words, even if the mechanic finds that the aircraft isn't airworthy, they're still required to make and sign a logbook entry to document that. The key point that Mike Busch makes is that what happens next is up to the owner, not the mechanic:

The owner must correct those discrepancies before the aircraft may
flown. He is free to have them corrected by any mechanic he chooses.
There’s no need for the inspecting mechanic to look at the aircraft
again. The next annual comes due in 12 calendar months.

The owner might ask the same mechanic who did the inspection to make repairs, or a different mechanic at the same airport. Or they might ask the FAA for a ferry permit to move the aircraft to another location for maintenance. Or they might just fly the aircraft anyway, knowing it isn't airworthy.
As for other maintenance that isn't a required inspection, 43.9(a)(4) says (emphasis mine):

If the work performed on the aircraft, airframe, aircraft engine,
propeller, appliance, or component part has been performed
satisfactorily, the signature, certificate number, and kind of
certificate held by the person approving the work. The signature
constitutes the approval for return to service only for the work
performed.

Here's Mike Busch's commentary on that:

Those two sentences bear careful reading. The first sentence makes it
explicitly clear that a mechanic’s signature on a 43.9 logbook entry
does not signify anything about the airworthiness of the aircraft. It
signifies only that the work the mechanic completed was performed
satisfactorily. [...]
The second sentence makes it clear that the mechanic’s signature does
not constitute an approval for return to service for the entire
aircraft (the way an annual inspection signoff does). It constitutes
approval only for the work the mechanic performed.

So coming back to your specific wiring scenario, if the mechanic performs a required inspection and determines that the change makes the aircraft unairworthy, they would document it as a discrepancy per 43.11. That means the aircraft is not airworthy, but how to make it airworthy is your decision, not the mechanic's decision.
Finally, if you're not sure how a mechanic would react to your wiring change you should probably find out. It's possible that the aircraft is already unairworthy, but that depends on a lot of details: type of aircraft, certified vs experimental, any applicable ADs etc.

and of course after the fixes are done, another inspection is needed to certify the aircraft as airworthy. So it's best to have things you know to be out of spec fixed BEFORE you have the aircraft inspected.

I would say that adding a switch counts as a minor alteration and only needs an A&P's signature. If it isn't in the major alterations list in Appendix A, it's a minor by definition.  I don't think adding a switch meets the criteria.  The AI will see the change, go and check and see of an A&P made a log entry, and if not just inspect it and sign it off if satisfied with the work (and charging for it), or add it as a discrepancy if not.  The biggest problem here is the AI now knows the owner likes to do home projects on the sly and will be wondering what other surprises are hidden in the plane.

Terrible misinformation on here. The truth is YES an A&P automatically grounds a plane when and if he does not release it as airworthy. If he signs log book that the plane is not airworthy THEN It is illegal for any pilot or owner to fly any plane that is not airworthy. A plane with a log entry it is not airworthy is illegal to fly. It is effectively grounded. NOW if the A&P has not released the plane as airworthy the owner now must get FSDO permission to fly that plane in a ferry permit or have plane fixed where it is some way that is legal. . In addition, if for any reason the A&P suspects that the pilot or owner may fly the plane even after it is found not airworthy, he can contact the FSDO and have a PI do a ramp inspection and ground the plane permanently on the spot.  SO when can a A&P write a determination it is not airworthy? After any repair or maintenance where they have identified a non airworthy condition of the plane the owner does not want fixed. It must then be recorded in log book MAKING the plane at that instant, grounded for all intense purposes, since it is illegal for anyone to fly it without being airworthy. The owner pilot will need FSDO permission to ferry or fix plane another way that is legal . NOW you have the real truth. It is NOT up to the pilot if he wants to fly a plane a A&P has determined is not airworthy. IT is up to FSDO and it is illegal to fly unless they give a ferry permit, or unless the problem has been corrected and another A&P signs off it has been corrected, followed by a I/A return to service. Because only a I/A can sign off any discrepancy that has been repaired as passing inspection and able to return to service, AFTER it was found to have been not airworthy by a A&P. Another A&P repair alone can not return a plane to service after a non airworthy discrepancy has been noted in the log by a A&P. A&P can only return to service 100hr inspections and regular maintenance of planes that are still airworthy, withing the annual time period of a planes Last 1 yr airworthiness authorization. by doing so ALL A&Ps do is prove a plane is still airworthy within the annual period. . They can not return to service any repaired discrepancies on a plane that is already documented as not airworthy. Once a plane is stated as not airworthy, only a I/A can return to service, or a DER or the FAA. Period.

Hi Joseph, and welcome to Aviation.SE. What you've posted is a bit of a wall of text; please add paragraph breaks to make it more easily readable.  Also, you make a lot of statements, but the current text doesn't provide any sources or references, making it all "somebody on the internet said that..." at this point. While I suspect that most or all of your statements can be supported by the applicable FARs, somebody who's looking for an authoritative answer doesn't have much to work from at this point from your post. It would help if you add links to the relevant regulations.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.


text:
How can I be US aircraft maintenance technician?
I have only an airframe-related experience of 8 years without a license, but have logbook, my US citizenship in process, and want to become an aircraft technician. How can I become a US aircraft maintenance technician? I have Pakistani aviation experience in airframe/B1.

The FAA has a guide to becoming an aviation mechanic, and the related regulations are in 14 CFR 65 Subpart D. In summary:

Be at least 18 and be able to speak and write English
Have 18 months experience working on airframes or powerplants; or, 30 months experience working on both
Pass three tests: written, oral and practical

The experience requirements are in 65.77 and it doesn't say that you need to have worked in the US. So your experience in Pakistan should be valid, although I have no idea what documentation you need to provide.
If you're inside the US, there's no requirement to be a US citizen, although to work and/or study in the US you need an appropriate visa (or green card). Some employers might require citizenship for specific positions, for security or other reasons.
If you're outside the US, you have to be a citizen or permanent resident, or get special approval per 65.3:

§65.3   Certification of foreign airmen other than flight crewmembers.
A person who is neither a U.S. citizen nor a resident alien is issued
a certificate under subpart D of this part, outside the United States,
only when the Administrator finds that the certificate is needed for
the operation or continued airworthiness of a U.S.-registered civil
aircraft.


text:
What is scrub drag?
I've come across the term "scrub drag" in several papers recently, but can't find a detailed definition of this phenomenon.  This is specifically in reference to aircraft and seems to apply to engine positioning.
For example: "Coe
observes that scrub drag in cruise could be reduced by pylon mounting the engine
over the wing", from the first paper below.
The term is assumed to be known to the reader in the following papers:
A Computational Study of Engine Deflection Using a Circulation Control Wing
Conceptual Design of a Next Generation, 150 Passenger Commercial Transport
A Study on Aerodynamic Design Integration of the Supercritical Wing and Rear-Mounted Engine Configuration
The same term in the context of automotive design does not appear to apply.

I'm gonna go with, drag that can't get no love from me.

This paper seems to show it's drag caused by air disturbed by a front mounted propeller.
Page 9:

Single Pusher: This configuration reduces scrub drag as the aircraft operates in clean air. However, a motor mounted aft of the aircraft will operate in the wing/fuselage wake, which causes a decrease in propulsion efficiency.

https://www.aiaadbf.org/2016TopReports.aspx

That interpretation doesn't seem to match up with other statements, especially ones dealing with turbofan powered aircraft (no propellors), or with there being a difference in cruise mode as opposed to other modes of operation (engine position remains the same).  It does seem to have something to do with turbulent flow.

It was the only reference I could find to scrub drag.


text:
How long would it take the Boeing 787 to go around the world?
How long would it take the B787 to complete a full trip around the world and how many times would they have to refuel in that trip?

What do you mean by "going around the world"? Great circle distance around the equator or what?

There are several definition of circumnavigation. You should edit your question to make it more specific. For example, make at least 40000km, land at your forst take-oof airport, pass through all meridians,....

Its not so simple, that type of flight would have to plan its refueling stops for airports that could handle that type of aircraft. It wouldn't fly in a straight line. You could just divide out the range by the fuel it could carry and get an X miles per gallon and Y segments, but it will be more.

Even in a no wind situation the earth is NOT a sphere it is wider at the equator than it is along the meridian, this will have a tangible effect on the outcome, which way is this plane going?

Related question on astronomy, with Jupiter instead of Earth... in case that helps clarify what the question is asking for.

@fooot, This one and its comments may be of interest too.

This question talks about circumnavigation.

If you departed from an airport near the equator and flew around the world with an average ground speed of 480 knots it would take about 45 hours plus the time needed for 2 fuel stops.
Great Circle Mapper

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text:
Do modern jet engines need igniters?
I've read that modern aircraft don't need spark plugs like conventional aircraft do, as the jet fuel autoignites once it's mixed with the hot air exiting from the compressors. Then why do aircraft like the Airbus A320 still have 2 igniters in the combustion chamber, to ignite the air fuel mixture?

"as the jet fuel autoignites once it's mixed with the hot air exiting from the compressors" Where's the hot air when the plane is starting? Exactly. That's what the igniters are for.

@Mast the air is very hot after the compressor stage (due to the compression itself), so it is not an unreasonable assumption that it would autoignite. It's just that the starter is not capable of reaching enough compression to reach the autoignition temperature.

Oh, it's definitely warm. Hot enough not to touch it. But not autoignition-hot indeed. Should have been more precise, thank you @Bianfable

Jet fuel will not self-ignite when starting a modern turbine engine.
This article from the WingMag Aviation Magazine says:

As the temperature isn’t quite sufficient to initiate self-ignition (the autoignition temperature of aviation fuel is around 220 degrees Celsius), spark plugs are arranged around the combustion chamber. They generate a spark that ignites the air-fuel mixture and the turbine now drives the fan and compressor through a shaft, as described above. The exhaust gas temperatures begin to rise and the engine will now keep “running” as long as there is a supply of aviation fuel. The igniters are switched off by the FADEC once the exhaust gas reaches a certain temperature because they are no longer needed.

Once the engine is running at a sufficiently high speed, the autoignition temperature can be sustained (same article):

[The fuel] is injected into the combustion chamber through several “fuel nozzles” where it can self-ignite and continue to run if the temperatures are sufficiently high.

See also How is combustion flame maintained in the combustion chamber after igniters are switched off?
Igniters are not only needed for a cold start. They are also used in case of an engine flameout. From the Boeing 747-8 FCOM (7.20.8 - Engine System Description, emphasis mine):

Auto-Relight
An auto-relight capability is provided for flameout and rollback protection.
Whenever the EEC detects an engine flamout or rollback, both igniters are
activated. A flameout is detected when a rapid decrease in N2 occurs. A rollback
is detected when N2 falls below idle RPM or compressor pressure falls below idle
minimums.

Note that this is for a modern jet engine, the General Electric GEnx.

@Johnson if this answer was helpful, click the tick on the left. This will mark the answer as accepted, and reward the answerer for their effort.

Continuous ignition can also be used in cases where a flameout is especially likely (icing, very heavy rain, heavy turbulence) or especially undesirable (takeoff and landing).

@hobbs Yes, some aircraft (like e.g. 737 and 747-400) have a continuous ignition switch to provide a continuous (but lower) power to the igniters. The 747-8 however, does not have a continuous ignition switch any more. It relies only on the auto-relight function.

The autoignition temperature of kerosene is 428F.   The temperature coming out of the last stage is usually above that at moderate power settings, up to 7-800 F or more at takeoff thrust.  However the temperature has dropped somewhat by the time the air gets to the fuel nozzle, and in heavy rain there may be superheated water cooling things down some more, and also when the engine flames out it cools off rapidly as the compressor rpm decays. And during start the discharge temperature doesn't get hot enough until after it's running.
So for starting at any temperature you need the igniters, and for relights if there has been any time to cool down you need the igniters.  Theoretically, at moderate to high power settings, if the flame went out it would auto-relight immediately if the fuel was still there or was immediately restored (you'd get a big ITT/TIT spike), but what normally happens is it flames out, say due to rain or something, and things cool off enough that the igniters are needed to get it going again.
Or at least there is a high probability that the igniters will be needed to light off.  So pretty much all modern engines have igniters that come on automatically following a flameout, and if not, you are supposed to operate with continuous ignition in heavy rain and extreme turbulence.

those are to help the fuel ignite during a cold start, when all the components in the hot section are at ambient temperature and there exist no flames in the burner cans. And as pointed out by Jan Hudec, the start motor (if the engine has one) or the APU (if it does not) cannot spin the compressor fast enough to heat the air to the point where the injected fuel will autoignite.

More importantly, the starter can't achieve that compression ratio.

@jan, exactly, will edit my answer. -NN

Most modern jet engines don't have a starter motor.  They use engine (or APU or huffer) bleed air to spin the engine up.

will edit......

The easiest way to understand this is by comparison to a piston 4 cycle engine:  In the piston engine the 4 cycles occur in the same location, but at different times.  I.e. the combustion cycle gives way to the exhaust cycle as the fuel is burned up and pushed out the exhaust valve.  On each subsequent cycle the fuel air mixture must be ignited again.
Contrast this with a turbine engine where each of the 4 cycles occurs simultaneously, but at different locations in the engine.  In other words, the burner section is burning continuously and the igniters are only needed to light it at the start, or to relight if the flame blows out.
Think of it like a propane or natural gas stove or torch:  You need a spark to light the flame initially, but the need for a spark goes away as long as the flame stays burning.

Well, a piston 4-cycle engine with compression ratio comparable to typical turbine is a compression-ignition one and does not have any spark plugs at all.

Are you talking about a diesel engine?  Sure, I agree with you then.  And I don't think you are refuting any of my points...


text:
Can computer imposed inputs be overridden on the Boeing 737-MAX?
During Lion Air Flight 610, a Boeing 737-MAX, it seems that the angle of attack sensors caused the software to believe the aircraft was stalling, thus initiating a nose-down pitch to recover from the (non-existent) stall.
I was under the impression that the difference between Airbus' fly-by-wire system and the Boeing types, is that Boeing aircraft can have their computer imposed inputs overridden (as discussed in my previous question).
So in the case of this flight, would pulling up have not recovered the flight, and if so, why not?

Also, there are many variants of the Boeing 737, presumably with different behaviors in specific situations. It would be better if you specify which variant you have in mind.

@aCVn The 737 does not have a stick pusher, only a stick shaker. Usually, only T-tail aircraft use stick pushers because they are more susceptible to deep stalls.

I mean the MAX @aCVn , since that's the incident I referenced. Also, it seems that it adjusted the trim to achieve nose down... please read the report.

@Bianfable I didn't know that. Thanks for the correction.

@Cloud We shouldn't need to refer to external material just to know what you're asking. Sure, the information is out there, but it's better to include pertinent information in the question itself. Basically, by making it as easy as possible for people to answer, you're more likely to get (good) answers.

An independent analysis of published flight data plots for that accident suggested that the column force plots allegedly reveal that on this flight the pitch disconnect (intended to retain controllability for one crew member in case the other control column jammed) was triggered, rendering the pilot flying in control of one elevator panel only (the other elevator could have been controlled by the other crew member) which then was insufficient to counter stabilizer nose down moment at full aft deflection of a single control column.

The new 737 MAX has a more advanced pitch control system called maneuvering characteristics augmentation system (MCAS), which automatically adjusts the stabilator trim in case of high thrust and high angle of attack (AOA). This makes it very hard to overcome the down force with elevator input alone (pulling up).
In case of a wrong trim adjustment (e.g. due to an erroneous AOA sensor), the pilots have to adjust the trim manually and maybe even turn off the stab trim via the cutout switches in the center pedestal (shown below). This was apparently not conveyed properly in the MAX differences training and the FAA has since issued an Emergency Airworthiness Directive.
737 NG:

(image source: flaps2approach.com)
737 MAX:

(image source: feitoffake.wordpress.com)
So compared to an Airbus, the Boeing pilots can still have all manual control, but they have to use manual pitch trim.

"The new 737 MAX has a more advanced stall protection system called maneuvering characteristics augmentation system (MCAS)" — from what I've read, MCAS has nothing to do with stall protection. It is meant to add more counterforce to the column at high AOA to comply with certification requirements. "So compared to an Airbus, the Boeing pilots can still have all manual control" — compared to the NG, the MAX pilots lose capability to turn off STS/MCAS but to continue using electric trim. The image you attached does not apply to the MAX.

@RustyCore I agree, the term "stall protection system" is not quite correct. MCAS was added to make the MAX feel like the NG in high thrust and high AOA situations, not to prevent stalls. The image I added does however apply to the MAX (even though this one is from an NG), which has the same cutout switches.

Um, no. The MAX has different switches: https://aviation.stackexchange.com/questions/64442

@RustyCore The switches are identical (labeled differently) and turning both off (which is what the Runaway stabilizer checklist from the QRH says anyway: "4) If the runaway continues: STAB TRIM CUTOUT switches (both) ... CUTOUT") also results in identical behavior in both the NG and the MAX. Only turning one of them off would result in different behavior. I added a second image.

"turning one of them off would result in different behavior" — which means they are different. Different labeling, different function, very different wiring if the schematics found elsewhere is to be believed. Turning any one of the switches on the MAX is like turning both of them on the NG, and turning off STS/MCAS while retaining electric trim is not possible anymore like it was on the NG. Boeing added a very powerful automatic feature and at the same time removed a major tool to defeat this feature.

So in the case of this flight, would pulling up have not recovered the flight?

"Pulling up" (pulling back on the yoke) would NOT have recovered the flight. In fact, the Lion Air pilots were pulling back on the yoke for most of the flight (see the flight data recorder's Control Column Force graph).

Can computer imposed inputs be overridden on the Boeing 737-MAX?

Yes. Pulling back on the yoke controls the elevator and would override any computer control of the elevator. The problem with the Lion Air flight is that the MCAS (which kicked in due to erroneous AoA sensor data) controls the stabilizer, not the elevator. That's why pulling back on the yoke didn't work.
The stabilizer is the entire "tail wing", whereas the elevator is just the flap on the trailing edge of the stabilizer. When the stabilizer is at full deflection, its effect on pitch will be greater than the elevator (yoke).
The stabilizer trim can easily be overridden as well (but not with the yoke). You can either:

Use the thumb switch to control the stabilizer trim electrically.
Hit the Stabilizer Trim Cutout switches (that would disable the automatic Stabilizer Trim, including MCAS)
Manually/physically turn the Stabilizer Trim Wheel.

The problem with the Lion Air flight is that they probably did not recognize that it was the trim they needed to override, so pulling back on the yoke was futile.

I am confused again. This picture shows ailerons on the main wing and elevators on the tailplane: http://www.dutchops.com/Portfolio_Marcel/Articles/Flight%20Controls/B737_Flight_Controls/B737_Primary_Flight_Controls.html So, (1) does the whole stabilizer moves on the 737, (2) or only rearmost parts, elevators? (3) Or both? (4) Or elevators are installed on the main wing, along with ailerons? Ah, seems (3) is correct: "The elevator is used for pitching the aircraft and is attached to the horizontal stabilizer, which can also be moved by systems like stabilizer trim and autopilot trim."

So, when Boeing says that operation of MCAS will "likely go unnoticed by the pilot," they are lying? Because when MCAS is operating, the trim wheels are rotating? This cannot be "unnoticed".

@RustyCore, Yes, the trim wheels are moving during MCAS activity, but the trim wheels are around the knees. So if a pilot is looking out the window or at some other instrument, they may not notice the trim wheel moving. Also, STS moves the trim wheel, so the pilot may assume STS is causing the trim wheel activity and not MCAS.

Looking at the videos, the stabilizer trim wheels make such a loud noise, that one has to be completely out to not notice their rotation (I haven't been in a pilot's cabin so cannot judge). Both STS and MCAS causing stabilizer adjustment, so if a pilot notices something funky, it is not really important which subsystem caused the funkiness, because the operations to turn on automatics are the same (stab trim switch + manual crank), and it is not that specific subsystem is turned off. It seems that STS uses the same sensors as MCAS, although I don't know whether it uses both AOA sensors or one.

@RustyCore The stickshaker (which was going off the entire time) is much louder than the trim wheels. STS also uses only one AOA sensor.

Mike, the 737NG FCOM says that STS uses "inputs of stabilizer position, thrust lever position, airspeed and vertical speed". No mention of angle of attack. Where do you get your information from? Do you have a link?

How would the crew of JT610 not recognize the trim as the issue? The Captain kept them in-trim through 21 MCAS activations. The F/O, to whom he transferred control to, did not.

The elevator DOES counteract the pitch change caused by MCAS operating erroneously, albeit with increasing aerodynamic resistance and with limited effectiveness compared with the capability of the stabilizer. The crew DID recognize the trim as the issue and used both elevator and electric trim intuituvely, as expected, to interrupt MCAS and restore pitch to the desired state each time MCAS activated. The indications are that they simply didn't know what to do beyond that. It appears they had never even heard of a stab trim runaway, let alone done the training for it.

Mike's answer is right. The yoke controls the elevator not the stabilizer.
MCAS trims the stabilizer, which can be overriden through the thumb switch, trim cutout switches (disables MCAS and the thumb switch), or by manually turning the Trim Wheel. (To see what that looks like, refer to the end of this video.)
To expand on Mike's answer, the preliminary report for the Ethiopian Airlines flight has been published.
According to the report, after MCAS activated, the pilots first executed option 2 (hit the Stabilizer Trim Cutout switches). They then attempted to trim manually via the wheel (option 3), which was apparently unsuccessful - this can be due to the aerodynamic forces on the stabilizer making the wheel very hard to turn. At last, they reengaged electric trim, and trimmed in the Aircraft Nose Up (ANU) direction with the thumb switch (option 1).
A few seconds later, MCAS reengaged in the Aircraft Nose Down (AND) direction for 5 seconds. 15 seconds later, the recording ends.
According to the report, they were pulling on the yoke continuously to keep level flight during most of the recording.

No, the Lion Air crew successfully undid the MCAS pitch change 22 times, using elevator and electric pitch trim; altitude not a factor as MCAS only activates when flaps are fully retracted. When MCAS activated the last two times, control had been handed over to the FO who did nothing to counter the runaway trim other than simply pulling on the elevator control yoke with increasing force until it exceeded his ability. The Captain, who had been countering MCAS, now had his head buried in a manual and dismissed the FO complaint the controls were too heavy. The fatal dive: due to pilot inaction.

The best is to refer to the following website:
https://www.boeing.com/commercial/737max/737-max-software-updates.page
You may read the following in the introduction:

 The Maneuvering Characteristics Augmentation System (MCAS) flight control law was designed and certified for the 737 MAX to enhance the pitch stability of the airplane – so that it feels and flies like other 737s.
  MCAS is designed to activate in manual flight, with the airplane’s flaps up, at an elevated Angle of Attack (AOA).
  Boeing has developed an MCAS software update to provide additional layers of protection if the AOA sensors provide erroneous data. The software was put through hundreds of hours of analysis, laboratory testing, verification in a simulator and two test flights, including an in-flight certification test with Federal Aviation Administration (FAA) representatives on board as observers.

To answer your question more precisely, please note pilots are trained to follow procedures. For every failure there is a corresponding checklist. In case of failures requiring urgent response, the pilots will perform the most important actions by memory and will then follow the checklist. On the 737 MAX it is a paper checklist and not an electronic checklist that is immediately displayed like on the Boeing 777.
Concerning the trim behavior related to MCAS, no clear checklist was available because the MCAS function was not available in the FCOM.
Normally, in case of runaway stabilizer trim the pilot may instinctively move the column in the opposite direction of the runaway. The next action is to switch off the cutout switches in order to be able to crank the wheel manually, however if the aircraft speed is very high and the angular deflection of the trim is high it becomes very hard to crank the wheel.

It is not merely the lack of a checklist. Pilots of the EA302 did have updated checklist and turned stab trim switches off. They could not adjust the stab manually, so they turned the switches back on, and MCAS flew them into the ground. Supposedly, the updated MCAS is less aggressive, but there is nothing in the blurb about changed functionality of the stab trim cutout switches, meaning that similarly to the EA302 flight, pilots still cannot turn off MCAS while keeping electric trim. The situation will be less likely to happen, but if it happens it will be as deadly as it was for the EA302.

With the modification in case of AOA disagree the MCAS autotrim is inhibited, so we get into a degraded mode which is not very safe; it would  be better to add a third AOA sensor in order to be able to isolate a faulty one.

There was no procedure for shutting off MCAS and there still isn't one, because MCAS cannot be shutoff. The procedure for a runaway stab trim used to have an initial troubleshooting section to determine whether it was caused by the autopilot or some other electrical fault. Accordingly, the pilits could break the autopilot's control lines to the stabilizer and keep electric trim operational. This proved to waste too much time and pilots couldn't remember which switch did what, so Boeing simplified things… simply shut down the electric trim system altogether; either cutout switch will do it.


text:
SID revisions numbers
I have noticed most instrument departures have a number after them. TRUKN2 GAP7 CITY3 for example at KSFO. Do the numbers represent a revision of say TRUNK1 GAP6 CITY3 for maybe a change in and altitude due to construction of a building, or FAA SID Departure criteria? Thanks in advance!

Yes, that is correct. The number indicates a version of the procedure, and runs from 1 to 9 before cycling back to 1. The changes may be due to newly-surveyed obstructions, or changes in ATC procedures to allow for more efficient traffic flow, or the decommissioning of conventional NAVAIDs (VORs) which used to be part of a procedure.
The FAA publishes documents highlighting the changes between successive versions of procedures, for example this one shows the changes between the FROGZ TWO and FROGZ THREE arrivals. These and other terminal procedures can be accessed at the Terminal Procedures site. Third parties like SkyVector also show these documents.

Wonderful answer and explanation ! Thank you so much for you help and sharing your knowledge!


text:
What is MCAS' purpose?
In a recent question, one of the answers states:

MCAS has nothing whatsoever to do with stall recovery or stall behaviour.

It was implemented strictly to make the MAX's pitch stick force response in certain regimes match older 737s, to avoid an additional training requirement necessitating Sim time.

But after reading this article, I'm confused about what exactly MCAS was intended for, as I thought it was stall recovery.

This website has a very comprehensive description and analysis of MCAS: http://www.b737.org.uk/mcas.htm
I find it to be the most accurate of all the various sources often quoted.
Here is a small quote from that website:

MCAS is a longitudinal stability enhancement. It is not for stall prevention (although indirectly it helps) or to make the MAX handle like the NG (although it does); it was introduced to counteract the non-linear lift generated by the LEAP-1B engine nacelles at high AoA and give a steady increase in stick force as the stall is approached as required by regulation.
The LEAP engine nacelles are larger and had to be mounted slightly higher and further forward from the previous NG CFM56-7 engines to give the necessary ground clearance. This new location and larger size of nacelle cause the vortex flow off the nacelle body to produce lift at high AoA. As the nacelle is ahead of the C of G, this lift causes a slight pitch-up effect (ie a reducing stick force) which could lead the pilot to inadvertently pull the yoke further aft than intended bringing the aircraft closer towards the stall. This abnormal nose-up pitching is not allowable under 14CFR §25.203(a) "Stall characteristics". Several aerodynamic solutions were introduced such as revising the leading edge stall strip and modifying the leading edge vortilons but they were insufficient to pass regulation. MCAS was therefore introduced to give an automatic nose down stabilizer input during elevated AoA when flaps are up.

Don't go by Wikipedia. Read the congressional report on the issue here. Stall comes into the issue in a peripheral way, but it's not what the system itself was addressing. It was addressing behaviour that could nudge the unwary pilot closer to the stall region you might say, but not stall recovery, or behaviour in the stall itself (maybe "nothing whatsoever" is a bit too strong, but it was a secondary issue at minimum).
The main issue was, while doing turns while flaps up and at fairly low speed, the force required to hold the nose attitude while in the turn would go down in a subtle way, and a pilot holding, say, 15lbs of back pressure to maintain the pitch attitude in the turn would notice the nose drifting up without increasing back pressure. Normally, someone paying attention while they were flying would notice this and react with a relaxation of back pressure instinctively, but it was an odd characteristic nonetheless, certainly different from older 737s.  The other large issue was the pitch up with thrust was more pronounced than with with older 737s, a result of the same factors.
These sorts of software band aids are not that unusual.  The 747-8 actually uses the FBW ailerons, which are normally inactive at high speed, to work as active oscillation dampers to stop the wing tips from shaking up and down at high speed. And whenever you see vortex generators stuck on a wing or tail somewhere, they are usually a physical band aid to fix something found during testing.
The scandal about the MAX is mostly in Boeing's implementation of the band aid.

“They might have been able to just train pilots to recognize this behaviour and probably certified it as is, but at minimum it would have forced airlines to implement a simulator session to make sure pilots recognized the condition and could respond to it.”  I disagree. The FAA Regs would not allow it to be certified without MCAS. You can’t make exceptions and then train pilots to fly an aircraft that does not behave the way aircraft normally behave. It would not be safe.

I'll take that para out.  My understanding was that the behaviour wasn't bad enough to be "uncertifiable" in absolute terms, just uncertifiable without adding a sim training requirement and MCAS was intended to replicate the older 737 more precisely to avoid that.  But perhaps I had the wrong idea.

There has been lots of speculation, and lots of misunderstanding, by all kinds of “experts”. I have never seen any actual documentation to say MCAS was ONLY needed for fleet Type Rating commonality. To me, it would have to be one reason or the other, not a little bit of both.

I'm hearing now that the final go forward with the software/wiring changes to recertify the system will include special sim training anyway, so that ship has sailed for good and all that grief was for naught.  I used to joke at my company that there needed to be a "False Economy Committee" of ppl with huge brains and special powers who can predict the ultimate cost of attempts to cut corners.  Usually it was related to choosing supplier X who's junk parts are cheaper, and ultimately cost 3 times as much in the long run.  This is on a whole new level.


text:
How much more dangerous is it to fly in Mexico than Europe?
Recently, Mexico's air safety rating was downgraded. How much more dangerous (in terms of statistical chances of fatal accident) is it to travel with a Mexican airline, than a European one?

If you mean flying as a passenger on a commercial flight, your question really belongs on the Travel site.

Why comparing with Europe? It is FAA who downgraded, not EASA.

@jamesqf Disagree, aviation safety and aviation regulations are both explicitly on topic here. Questions may be on topic on multiple sites and that's OK.

@user3067860: If the OP was asking as a pilot, it would be relevant.  As a passenger, it's not.  If it's on topic, it shouldn't be.

@jamesqf It specifically says that even aviation enthusiasts are allowed to ask questions. What would really change, though, if the question started off with "I am just about to start pilot training in Mexico next week..." and then continued exactly as written... Would it be on topic? Would the answer change at all? If the answers are "yes" and "no" in that order, then why discriminate based on who's hand was on the keyboard?

@user3067860 jamesqf has a hobby... closing questions.

To answer your title question with some numbers, here are the accident rates (number of accidents per million departures) for the world, Europe and Mexico from 2008 to 2020:

(Data from ICAO - Accident Statistics)
It's not quite what you asked about since the ICAO data does not filter by airline origin country and the accidents might include non-fatal ones (but not incidents, see What is the difference between aviation "accident" and "incident"?).
Nonetheless, it's clear that Mexico is not significantly less safe than Europe. The average accident rate since 2008 was 3.27 in Europe, but only 2.06 in Mexico. The plot above should make it clear that these numbers are fluctuating quite a lot because of the small number of accidents in each year. For example, the large peak for Mexico in 2010 is the result of only 6 accidents. One should therefore be careful when deriving any conclusions from these numbers.

It's a good idea to provide these stats, but be aware the State criteria doesn't count accidents by operator State but by accident location. E.g in France stats Air France AF447 in 2009 (over east of Brazil) isn't counted while Germanwings 4U9525 in 2015 (over France) is counted.

@mins Yes, that's what I meant by "the ICAO data does not filter by airline origin country". Unfortunately, I couldn't find such a dataset. I found some with number of accidents by airline, but without a suitable denominator to get a rate, such a dataset is useless.

Not disputing anything included in this answer, but it is very important to understand that reading stats and interpreting graphs is a whole science of it's own. To draw any conclusions, one has to be familiar with the subject in question, as well as how the stats were derived from data. Only thing that can be safely stated based on the graph above is that apart from 2010, flying is generally about as safe in Mexico as in the rest of the world.

...there is solid truth in the saying "lies, damned lies, and statistics"...

@Jpe61 Exactly, hence my last sentence. I would even go as far as saying in 2010 Mexico was just as safe as in the other years. It was just a statistical fluctuation.

@Jpe61: "Only thing that can be safely stated based on the graph above is that apart from 2010, flying is generally about as safe in Mexico as in the rest of the world" From what I see Mexico has been at least 2x or 3x better than the rest of the world in the last 10 years, in term of accidents over their territory, but maybe all accidents with Mexican airlines took places elsewhere, therefore decreasing the number of accidents over Mexico and increasing it for the rest of the world...

Exactly @mins really hard to say pretty much anything, as there are only a couple of accidents annually. Any occurence carries a lot of "weight". Considering the graph is Mexico vs. Europe vs rest of the world, you'd expect the rest of the world be a lot higher, as the rest of the world includes states with, well, sub par safety culture. Also, accidents do not necessarily relate to local safety culture at all, as they may be simple technical failures, or out-of-region-crew errors etc. etc...

If there's anything that following US passenger rail has taught me, it's that the sample period has to be appropriate for the number of departures.  Amtrak and the airlines used to trade safety crowns ... Amtrak was on top until Big Bayou Canot, then airlines were on top until 9/11.  All this is inaccurate, and a very silly game, because the sample sizes are way too small and are whipsaw'd by single events.

Short answer
Such audits are done by FAA to evaluate whether safety is correctly managed by the Mexican regulator in order to inform US DoT.
FAA found there is a problem in how Mexico is organized to supervise Mexican airlines, and consequently US DoT will prevent new Mexican carriers to enter the US market, freeze new routes creation, and will prohibit US carriers to sell seats on Mexican carrier aircraft, until this is fixed.
Nothing can really be drawn from the result. There are different domains evaluated, not being compliant in one domain is sufficient to be disqualified. FAA doesn't disclose the details of the non-compliance.
While this audit is performed by US for their own benefit, the framework used is the international framework of ICAO. ICAO also conducts audits in a continuous way. ICAO audit results can be viewed online with some details. Mexico was audited by ICAO in 2012. A comparison of results for Mexico (2012), US (2007) and France (2020, there is no "Europe" rating, ratings are by countries):

This comparison shows the Mexican authority is below the mean level regarding it's organization but also better than the US for licensing, legislation and operations.
How these categories are evaluated is described in the audit framework  ICAO Doc 9734 part A: The Establishment and Management of a State Safety Oversight System (respectively §3.3, §3.6, §3.1 and §3.2). Exactly like ISO 9000 audits, both ICAO and FAA audits are related to the management system ability to detect and correct problems in carrier operations, they are not related to how airline actually perform their operations.
Bottom line: We don't know how Mexican carriers perform, we know they are overseen by a non-compliant regulator according to FAA, due to one or more unknown weaknesses.
Details related to obligations of ICAO members and audits follow for those interested.

Details
Chicago Convention: Obligations of the States
Mexico is a member of ICAO and signatory of the Chicago Convention. As such Mexico has legal obligations, which fulfillment is delegated by Mexico government to AFAC, the Mexican regulatory agency, counterpart of FAA for Mexico. Contracting States have safety-related obligations, in particular with art. 37 and 12:

(37): Each contracting State undertakes to collaborate in securing the highest practical degree of uniformity in regulations, standards, procedures and organization in relation to aircraft, personnel, airways and auxiliary services in all matters in which such uniformity will facilitate and improve air navigation.

(12): Each contracting State undertakes to adopt measures to insure that every aircraft flying over or maneuvering within its territory and that every aircraft carrying its nationality mark, wherever such aircraft may be, shall comply with the rules and regulations relating to the flight and maneuver of aircraft there in force.

Each contracting State undertakes to keep its own regulations in these respects uniform, to the greatest possible extent, with those established from time to time under this Convention [...] Each contracting State undertakes to insure the prosecution of all persons violating the regulations applicable.

ICAO audit: USOAP-CMA
ICAO members agreed on a continuous audit approach which abbreviation is USOAP-CMA (USOAP website) to evaluate how States fulfill their obligations. ICAO and FAA audits are based on the same reference, ICAO Doc 9734 part A: The Establishment and Management of a State Safety Oversight System, which includes 8 domains, either for rule-making or for rule-enforcement:

Audited domains under USOAP-CMA and IASA, source ICAO Doc 9734
FAA and IASA
FAA conducted their IASA in May 2021. Objectives, from FAA:

IASA assessments determine compliance with these international Standards by focusing on the eight critical elements of an effective aviation safety oversight authority specified in ICAO Document 9734, Safety Oversight Manual.

Those eight critical elements include primary aviation legislation; specific operating regulations; State civil aviation system and safety oversight functions; technical personnel qualification and training; technical guidance, tools and the provision of safety critical information; licensing, certification, authorization, and approval obligations; surveillance obligations; and resolution of safety concerns.

The results are the one you're talking about in the question. The fact that Mexico was found to be not compliant to ICAO standards, means they were found to have one or more of these problems (source):

Lack of advisory documentation;
Shortage of experienced airworthiness staff;
Lack of control on important airworthiness related items such as issuance and enforcement of Airworthiness Directives, Minimum Equipment Lists, investigation of Service Difficulty Reports, etc.;
Lack of adequate technical data;
Absence of Air Operator Certification (AOC) systems,
Nonconformance to the requirements of the AOC System
Lack or shortage of adequately trained flight operations inspectors including a lack of type ratings;
Lack of updated company manuals for the use by airmen;
Inadequate proficiency check procedures;
Inadequately trained cabin attendants.

Consequences from a FAA standpoint
From FAA:

While the new rating allows Mexican air carriers to continue existing service to the United States, it prohibits any new service and routes. U.S. airlines will no longer be able to market and sell tickets with their names and designator codes on Mexican-operated flights. The FAA will increase its scrutiny of Mexican airline flights to the United States.

Re-occurence

It is the second time that Mexico is in Category 2. The first time happened in 2010, and it took the government four months to regain its prior status.

Source

It's not just that new Mexican carriers won't be allowed, it's also that existing ones can't open new routes to the U.S. and, perhaps more importantly, U.S. carriers can't sell codeshares onto the existing routes operated by Mexican carriers. This would be a pretty big deal for Delta, for example, which not only has Aeromexico as a close SkyTeam alliance partner, but actually owns a 49% stake in it. So, for example, they can't sell DL8XXX tickets on ATL-MEX flights operated by AM. Other Mexican carriers (e.g. Volaris, etc.) are also banned from opening new routes to the U.S.

@reirab: Good addition.

The airworthiness part is the most troubling, and clearly shows Mexico well below France and slightly below the U.S.

authors: Vladimir F Героям слава


text:
How can I land on an "IFR-only" airport?
I was flying Cessna Skyhawk in Microsoft Flight Simulator from Nowy Targ EPNT to Tatry LZTT. I set the weather to cloudy for the purpose of training (a little bit) IFR flight. In general quite big clouds could be spotted around heights of 3000-4000.
When I approached the target airport (and I assume MSFS should have "filed the flight plan" for me if it makes any sense), I asked for permission for landing with full stop, but in response I just got "Tatry airport is IFR only, request denied".
I tried to tune in to Warsaw air control and ask for IFR flight, but I was given order to "Raise to 13000 feet and maintain direction" while I was almost on approach to landing on LZTT.
How would such a situation be handled by the pilot and by ATC in reality? Note that it was well below 3000 and at that level visibility was very good (I mean good enough to do a perfectly safe VFR landing).
Please take into account (that's just informative), that even though I have some knowledge about IFR flying, contacting tower and general procedures, I'm a total amateur flying in the sim and learning about flying for sheer fun. I will be grateful for any explanation, which factors played role in that airport's tower refusing me landing there.

@Bianfable, a built-in ATC.

Pause the sim change the clouds and continue VFR.

@Bianfable, but that's actually what I want to know. How do I implement the proper procedure in the simulator is another story.

@Spook OK, I tried to rephrase the question to make it clear you're not asking about the sim. But I don't expect the sim ATC to work anything like reality (I haven't flown in MSFS, just my experience with older sim ATCs).

I don't see this as being answerable as the sim and reality don't match in a way that can be reconciled. Asking how you would go about an IFR flight in reality is very broad, IFR flight is an entire course with more than one book and hours of ground training to go along with it.

I’m voting to close this question because this is clearly a Gamimg question. In the real world ATC will not call an airport IFR and refuse an VFR aircraft to land when conditions are actually VFR.

@MikeSowsun, What requirements do you need to meet to land on a IFR airport?

@Spook LZTT is not an IFR airport. Your gaming simulation is wrong to call it IFR when the weather conditions are VFR.

Not familiar with MSFS, but did you correctly set the flight mode to IFR and define your flight plan when configuring your simulation? The select button under the aircraft type, once you have selected departure/arrival airports?

As I mention in another comment, LZTT is a restricted use airport that DOES, in fact, requires prior authorization before operating there.  Because flight sims strive for surprising ATC fidelity, the best way to represent this in the sim is to make it an “IFR airport.”  There are many cases of airports having a pre-approved clearance requirement, that are not technically an IFR requirement, but are still beyond “casual VFR” arrivals.

I think your direct question was "how would this be handled in real life?"  There are really three situations:

I am already flying on an approved IFR flight plan, and my approach clearance is just a continuation of that IFR flight plan.  The tower already knows I am coming, my aircraft type, and that I and the plane are IFR certified.

I need what is in the US called a "pop up" clearance.  I was NOT on an existing IFR flight plan, but ended up encountering unexpected IMC (instrument meteorological conditions).  This could be enroute, or this could happen just at the destination airport.  If both I, and my airplane, are certified and current for IFR, I request a pop-up clearance from the controller.  If approved, I will get an IFR clearance to continue.  But as you encountered in the sim, this could require mandatory resequencing and you might be shuffled around into a landing queue.

You declare an emergency.  If there is an urgent situation that in your judgment makes landing here, even below VFR minimums, a safety-of-flight issue, you can declare an emergency.  For instance, if weather is settling in all across the region rapidly, so proceeding to any other airport will be more dangerous, you'd declare an emergency.  If you are "urgent fuel" and can't safely make your way to another airport with visual conditions, you'd declare "urgent fuel".  Can't speak for other countries, but they'd (a) help you get down safely and in the US (b) almost certainly give you the phone number of the FSDO, where you'd set an appointment to be interviewed by a safety investigator.  After you landed, they'd say "123L, taxi to parking by taxiway bravo, advise ground when you are able to copy a phone number."

Those 3 scenarios assume the airport is experiencing IFR weather. The OP has clearly stated the airport was experiencing good VFR weather in this simulation.

@MikeSowsun Yes, but the OP also clearly stated that the airport was, for whatever reason, not accepting VFR arrivals.  I was not addressing the "why," which in real life would be very unlikely.  I was simply answering that, if an IFR approach is the only means of arrival, then, what could you do?  The options in VMC would be essentially the same--accept the denied clearance and fly elsewhere, pop-up IFR, or declare an emergency.  "Look, it's VMC, so I am landing anyway" probably won't work at a towered airport.  You'll still be copying a phone number for a different kind of call.

It's an obvious simulator glitch.  If the field is VMC in real life simply cancel IFR.

@MichaelHall I don’t see this as a simulator “glitch” in the traditional sense.  LZTT is a restricted use airport that requires “prior approval by the Aerodrome Operator” per AIP.  It’s not as simple as proceeding VFR.  Sim logic, which works surprisingly hard to mimic real ATC logic, would have no way to represent this other than designating the airport IFR only.  In the US there are also publicly owned, but restricted operation airports that fall outside our notion of Class B or D.

@Max, prior permission required (PPR) is separate from an ATC clearance.  In any case, I defer on any peculiarities of airports under the jurisdiction of EASA, and whether or not a flight simulator depicts them accurately.  Would you please elaborate on your last sentence though?

Anyway, the point of my comment is that the OP's scenario is the direct opposite of the solution in #2.  (Unless you need to be on an IFR clearances for purely administrative reasons)

@MichaelHall Example:  Military airports are quite often open to civilian/GA operations, but only with prior approval.  If you are a civilian and you have a meeting with an organization at Grand Forks AFB, you will almost vertically be granted an approval letter, where if you show up at the edge of the class D, the answer will almost certainly be no—the exception being declaring an emergency.  This isn’t reflected on sectionals, it just looks like a D on the map.  I do appreciate your comments and clarifications, thank you!

OK, I know that.  "Publicly owned" implied something else...  Either way, a pop-up IFR clearance won't satisfy PPR or "prior approval by the Aerodrome Operator".  #3 is the only real option if you simply must land there.


text:
What is the intent of 61.93(b)(1)?

(b) Authorization to perform certain solo flights and cross-country flights. A student pilot must obtain an endorsement from an authorized instructor to make solo flights from the airport where the student pilot normally receives training to another location. A student pilot who receives this endorsement must comply with the requirements of this paragraph.
(1) Solo flights may be made to another airport that is within 25 nautical miles from the airport where the student pilot normally receives training

The student must be endorsed with something along the lines of:

I certify that (First name, MI, Last name) has received the required training of section 61.93(b)(1). I have determined that he/she is proficient to practice solo takeoffs and landings at (airport name). The takeoffs and landings at (airport name) are subject to the following conditions: (List any applicable conditions or limitations.)

Emphasis on the word TO, in (1). I interpret this as "You must be endorsed to fly solo TO another airport for takeoffs and landings." However, my instructor, and it would seem the majority of the instructors at my school interpret it as "You must be endorsed to fly solo AT another airport for takeoffs and landings." The FARs seem to imply my interpretation, and the recommended endorsement seems to imply the opposite (in fact the endorsement doesn't even mention the 25nm part except in title).
The situation it is used in most commonly is the towered-field TO/LDG practice. There is a towered field about 10 miles north of our airport where we all go to practice. However, the instructor is required to fly dual up there, solo the student, and fly dual back. Is this endorsement required in that case?

I always add the limitation "Not without my knowledge and consent" as a fail-safe to my student pilot endorsements, so it probably wouldn't hurt to add this one just to be safe, but I'd like to know if it's necessary or not.

The intent of the regulation lies at the beginning, in 61,93(a):

§61.93   Solo cross-country flight requirements.
(a) General. (1) Except as provided in paragraph (b) of this section,
  a student pilot must meet the requirements of this section before—
(i) Conducting a solo cross-country flight, or any flight greater than
  25 nautical miles from the airport from where the flight originated.
(ii) Making a solo flight and landing at any location other than the
  airport of origination.

Training and an endorsement from an instructor is required before you fly cross-country or to another airport to land.  61.93(b) covers what the instructor must do with you before you can do anything specified in 61.93(a).
In the scenario that you describe you are not "Conducting a solo cross-country flight, or any flight greater than 25 nautical miles from the airport from where the flight originated" or "Making a solo flight and landing at any location other than the airport of origination" so subsection (b) does not need to be complied with at all.
The endorsement would only be needed if you were going to takeoff solo from your home airport, fly to the destination (which as you said is only 10 miles away), and land there since it is required by 61.93(a)(ii).